Divergence of $\int \limits_{1}^{+ \infty} \frac{x \sin(\log^2(x))}{\log(x+1)}dx$ I have improper integral:
$$\int \limits_{1}^{+ \infty} \frac{x \sin(\log^2(x))}{\log(x+1)}dx$$
I know, that it diverges, but I can't prove it. I tried to use the negation of Cauchy criterion for improper integrals, but I failed, because this integral doesn't have representation in elementary functions. So, how to prove the divergence?
 A: Let $F: [1,\infty) \to \Bbb{R}$ be defined by
$$ F(x) = \int_{1}^{x} \frac{t \sin(\log^2 t)}{\log(1+t)} \, dt. $$
The improper integral is then defined as the limit of $F(x)$ as $x\to\infty$ if it exists. So if the improper would have existed, then we would also have $F(e^{\sqrt{\smash[b]{\pi (n+1)}}}) - F(e^{\sqrt{\smash[b]{\pi n}}}) \to 0$ as $n\to\infty$. We show that this is not the case.
With the substitution $t = e^{\sqrt{x}}$, we find that
\begin{align*}
\left| F(e^{\sqrt{\smash[b]{\pi (n+1)}}}) - F(e^{\sqrt{\smash[b]{\pi n}}}) \right|
&= \left| \int_{\pi n}^{\pi(n+1)} \frac{e^{2\sqrt{x}} \sin x}{2\sqrt{x}\log(1+e^{\sqrt{x}})} \, dx \right| \\
&= \int_{0}^{\pi} \frac{e^{2\sqrt{x+n\pi}} \sin x}{2\sqrt{x+n\pi}\log(1+e^{\sqrt{x+n\pi}})} \, dx \\
&\geq \int_{0}^{\pi} \frac{e^{2\sqrt{n\pi}} \sin x}{2\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})} \, dx \\
&=\frac{e^{2\sqrt{n\pi}}}{\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})}.
\end{align*}
But notice that the denominator is $\sim n\pi$ as $n\to\infty$, while the numerator diverges much faster that polynomials. So
$$\lim_{n\to\infty} \frac{e^{2\sqrt{n\pi}}}{\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})} = \infty $$
and the desired conclusion follows.
